【NOI OpenJudge】【1.3】编程基础之算术表达式与顺序执行

01: A+B问题

#include
using namespace std;
int main(){
	int a, b;
	cin>>a>>b;
	cout<<a+b<<"\n";
	return 0;
}

02: 计算(a+b)*c的值

#include
using namespace std;
int main(){
	int a, b, c;
	cin>>a>>b>>c;
	cout<<(a+b)*c<<"\n";
	return 0;
}

03:计算(a+b)/c的值

#include
using namespace std;
int main(){
	int a, b, c;
	cin>>a>>b>>c;
	cout<<(a+b)/c<<"\n";
	return 0;
}

04: 带余除法

#include
using namespace std;
int main(){
	int a, b;
	cin>>a>>b;
	cout<<a/b<<' '<<a%b<<'\n';
	return 0;
}

05:计算分数的浮点数值

#include
#include
using namespace std;
int main(){
	double a, b;
	scanf("%lf%lf",&a,&b);
	double c = a/b;
	printf("%.9lf",c);
	return 0;
}

06:甲流疫情死亡率

#include
#include
using namespace std;
int main(){
	//int a, b;
	double a, b;
	scanf("%lf%lf",&a,&b);
	double c = b/a*100;
	printf("%.3lf%%",c);
	return 0;
}

07: 计算多项式的值

#include
#include
using namespace std;
int main(){
	double x, a, b, c, d;
	scanf("%lf%lf%lf%lf%lf",&x,&a,&b,&c,&d);
	double ans = x*x*x*a+b*x*x+c*x+d;
	printf("%.7lf",ans);
	return 0;
}

08:温度表达转化

#include
#include
using namespace std;
int main(){
	double x;
	scanf("%lf",&x);
	double ans = 5*(x-32)/9;
	printf("%.5lf",ans);
	return 0;
}

09:与圆相关的计算

#include
#include
using namespace std;
const double pi=3.14159;
int main(){
	double x;
	scanf("%lf",&x);
	printf("%.4lf %.4lf %.4lf",x*2,2*pi*x,pi*x*x);
	return 0;
}

10:计算并联电阻的阻值

#include
#include
using namespace std;
int main(){
	float a, b;
	scanf("%f%f",&a,&b);
	float r = 1/(1/a + 1/b);
	printf("%.2f",r);
	return 0;
}

11:计算浮点数相除的余数

#include
#include
using namespace std;
int main(){
	double a, b;
	scanf("%lf%lf",&a,&b);
	int k = 0;
	while(a-k*b >= 0){
		k++;
	}
	k--;
	printf("%g",a-k*b);
	return 0;
}

12: 计算球的体积

#include
#include
using namespace std;
int main(){
	double x;
	double pi = 3.14;
	scanf("%lf",&x);
	printf("%.2lf",4.0*pi*x*x*x/3);
	return 0;
}

13:反向输出一个三位数

#include
#include
using namespace std;
int cur;
void go(){
	char ch;
	if(++cur <= 3){
		ch = getchar();
		go();
		putchar(ch);
	}
}
int main(){
	go();
	return 0;
}

14:大象喝水

#include
#include
using namespace std;
int main(){
	double h, r;
	double pi = 3.14;
	scanf("%lf%lf",&h,&r);
	r*=0.01, h *= 0.01;
	double v = pi*r*r*h;
	double ans = 20/(v*1000);
	printf("%d",(int)(ans-0.0001)/1+1);
	return 0;
}

15:苹果和虫子

#include
#include
using namespace std;
int main(){
	int n, x, y;
	cin>>n>>x>>y;
	int b = (int)((y*1.0/x*1.0-0.001)+1);
	cout<<n-b<<'\n';
	return 0;
}

16:计算线段长度

#include
#include
#include
using namespace std;
int main(){
	double x1, y1, x2, y2;
	scanf("%lf%lf%lf%lf",&x1,&y1,&x2,&y2);
	printf("%.3lf\n",sqrt((x1-x2)*(x1-x2)+(y1-y2)*(y1-y2)));
	return 0;
}

17:计算三角形面积

#include
#include
#include
using namespace std;
int main(){
	double x1, y1, x2, y2, x3, y3;
	scanf("%lf%lf%lf%lf%lf%lf",&x1,&y1,&x2,&y2,&x3,&y3);
	double a = sqrt((x1-x2)*(x1-x2)+(y1-y2)*(y1-y2));
	double b = sqrt((x3-x2)*(x3-x2)+(y3-y2)*(y3-y2));
	double c = sqrt((x3-x1)*(x3-x1)+(y3-y1)*(y3-y1));
	double p = (a+b+c)/2;
	printf("%.2lf\n",sqrt(p*(p-a)*(p-b)*(p-c)));
	return 0;
}

18:等差数列末项计算

#include
#include
#include
using namespace std;
int main(){
	int a1, a2;
	cin>>a1>>a2;
	int d = a2-a1;
	int n;  cin>>n;
	cout<<a1+(n-1)*d<<'\n';
	return 0;
}

19:A*B问题

#include
using namespace std;
int main(){
	long long a, b;
	cin>>a>>b;
	cout<<a*b<<'\n';
	return 0;
}

20:计算2的幂

#include
using namespace std;
int main(){
	int n;  cin>>n;
	cout<<(1<<n);
	return 0;
}

转载于:https://www.cnblogs.com/gwj1314/p/10200060.html